Predict the major products when the following compounds are ir...

Question

Predict the major products when the following compounds are irradiated by light and treated with (1) 1 equivalent of Br₂ and (2) excess Br₂.

(a) isopropylbenzene

(b)

Accepted Answer

Hey everyone and welcome back into this video. We are going to propose the products formed by the light induced reaction of the following compounds with number equivalent of roaming and number two access of roaming we have two compounds, one of them is following. And the structure of the second compound is kevin Right now essentially we may start by thinking about the first part where we use only one equivalent of grooming. And we are going to look at Halloween. So for a the starting material is so Halloween meaning we are going to draw a method growth bonded to our Benzene ring, that's our starting material. And in this problem for the first part part, number one, we are treating Halloween with one equivalent of brahman. So we can say one equivalent of brahman is used notice that this reaction requires slight right? It's a slight induced reaction. So we will essentially include age knew which stands for light because we have a free radical hallucination. So because we're dealing with a free radical halogen nation, you may remember that bro mean is so selective. It will essentially react add the Benz Illich position. So we can say Ben's Illich position will be preferred. Now, even though it's not the case with Glory in roominess. Real original selective it wants to react at a site that can be resonance stabilized. Right, So Ben's Illich position. In this case if we look at Halloween would be this carbon atom because when we formed the radical in our reaction, its resonance stabilized and roaming prefers a stable radical. So that means if we're using one equivalent of Rome in one of the three hydrogen will be replaced with roaming. So the product for the first one would essentially be our benzene ring. Now instead of that metal group, we would end up with siege to right. We have to hide jen's left because one of the hydrogen atoms has been replaced with roaming. So once again we have identified our Ben's Illich position which is stabilized by the benzene ring When we formed the radical in our reaction at the site. Right. And then essentially it reacts with the blooming molecule to form our final product. Once again, we only Subsequute at one of the hydrogen atoms. Now in number two, the starting material is to Halloween once again. So we are going to draw Halloween. And the main difference is that for this reaction, we're not using one equivalent of roaming but instead we are using an excess amount of bro means So that would be access. Once again, we need to make sure that we're including light for this free radical halogen ation. And because we're using an excess amount of roaming all of the three high genes can be substituted. Right? So within the metal group there will be no more hydrogen is remaining. We will just get our final product which involves the presence of the benzene ring. And now the carbon atom here contains three bromo groups. So we're going to draw br another br and another one. So these are our products for this one. Now what about B? If we want to think about B we are going to apply the similar logic. We are going to draw our starting material so that we are benzene ring Which is bounded to a five member ring. Okay, now, as we previously said, roaming this regional selective, it wants to replace hydrogen atoms at the Benz Illich position. So the Benz Illich position is the first carbon atom that is bonded to the benzene ring. Notice that we have to Ben's Ilic positions here and here but due to the fact that there is this horizontal line of symmetry that is passing through our compound, the two positions are equivalent to each other. So that means we only have one unique Ben's Illich position and we are going to think about the top one. So for the first part where we use only one equivalent of roaming, we can say that we are dealing with part number one and we are adding one equivalent of bro mean we are going to replace one of the two hydrogen atoms here. Right, don't let's not forget to include light. So we are arriving light h nu below the arrow and now we are replacing one of the two hydrogen atoms here at the very top and we are getting our final product. So let's draw it out, that will be our benzene ring, let's draw it a bit clearer. So that would be our benzene ring and then we will draw our five member ring bonded. Sit and at the Benz Illich position we are going to add bro mean okay well done. So this is our product once again because we only use one equivalent, we have only replaced one of the two hydrogen atoms. And now what if we use access Now for number two we are going to redraw our starting material once again let's draw the reaction arrow. And now we are going to state that there's an excess amount of grooming. So we are writing booming inferences access below the are we will include the presence of light. And now this is really important because we have to Ben's Ilic positions right? We are going to replace all of the hydrogen atoms present at the Benz Ilic positions because we are using an excess amount of grooming so lets redraw the benzene ring Now our five member drink and because we have two hydrogen atoms at the top and two hydrogen atoms at the bottom, these two Bens Ilic positions will now contain two bromo substitue ints for the top men's Illich position and for the bottom end position we are going to add two bromo substitutes as well because our goal is to replace all of the hydrogen atoms at the Benz Ilic positions with bromo substitution. So that's why we have for bromine atoms added because there were four hydrogen atoms initially. So these are our products. We may essentially label these as our answer. So for a we have our products over here. So number 11 equivalent of roaming gives us the substitution of one hygiene within the metal group. While the excess of bromine replaces all the three high genes. And therefore be, we simply notice that one equivalent of roaming replaces one of the hydrogen atoms within only one position, right? Because they are equivalent to each other and we are only using one equivalent. That means only one hydrogen can be replaced. So we can either draw blooming on top or on the bottom. That's exactly the same thing. And for the excess amount of roaming, we are replacing all of the four hydrogen at the to Ben's Ilic positions. Now looking at the answer choices, we may simply conclude that the correct answer to this multiple choice question is a However, if you have any questions, don't have states, leave a comment down below in the comment section and thank you for watching

Predict the major products when the following compounds are irradiated by light and treated with (1) 1 equivalent of Br₂ and (2) excess Br₂.
(a) isopropylbenzene
(b)

Key Concepts

Free Radical Halogenation

Selectivity in Halogenation

Excess Halogen Effects

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